Transcript e - Mr. Schroeder

• Connecting two parallel plates to a battery produces
uniform electric field
e-
ep+
p+
p+
e-
e-
e
e
e-
p+
ee-
e-
 the electric field strength will depend on the distance
between the plates and the voltage of the battery

V
E 
d
E
V = voltage of battery
d = distance between the plates in m
= magnitude of electric field strength in V/m
V
J
N m N



m C m C m C
This formula only
applies to parallel
plates!
•electric field strength between the plates
is uniform (a charge placed at any point in
the field will experience the same force)
•the electric field outside of the plates is 0
•the magnitude of the electric field
depends on the magnitude of the charge
on each plate
•field lines point from the + plate to the plate
Examples
Two parallel plates are 3.00 cm apart and have a potential
difference of 7.10 x 103 V. If a charged object (q = 4.80 x
10-15 C) is placed in the field, what is the electric force
acting on it?

V
E 
d

7.10 x10 3 V
5
E 

2
.
3667
x
10
V /m
2
3.00 x10 m

E 
Fe
q

E q  Fe


( 2.3667 x10 5 N / C) 4.80 x10 15 C  1.14 x10 9 N
2. A pair of oppositely charged parallel plates are separated
by 5.33 mm and connected to a 600 V source of potential
difference.
a) What is the magnitude of the electric field between the
plates?
b) What is the magnitude of the force acting on an electron
placed anywhere between the plates?
c) How much work must be done on the electron to move it
to the negative plate if it is originally placed 2.90 mm
from the positive plate?